My orbitz problem

The gravitation force cancel out with the centrifugal force which equates into:

Gm1m2/r^2 = m2v^2/r where m1= mass of the sun, m2 = mass of planet

simplify the equation above I get: Gm1/r = v^2

This implies that if we move the plannet faster in tangent direction, the distance between 2 bodies must reduce to achieve a stable orbit. The opposite is true. Slowing down the planet will have to move further away from the sun to attain stable orbit.

Let's consider kepler's second law and conservation of momentum. Basically, the law said that when the planet is half radius of a reference position, the velocity increase with the factor of 2. But according the the equation above, the velocity should increase with the factor of 4. Who can shed some light into this?

This implies that if we move the plannet faster in tangent direction, the distance between 2 bodies must reduce to achieve a stable orbit. The opposite is true. Slowing down the planet will have to move further away from the sun to attain stable orbit.

Not following you …

Let's consider kepler's second law and conservation of momentum. Basically, the law said that when the planet is half radius of a reference position, the velocity increase with the factor of 2. But according the the equation above, the velocity should increase with the factor of 4. Who can shed some light into this?

ah … Kepler's second law and conservation of momentum apply to the same orbit.

They don't apply if you start shoving extra energy in!

If a planet is in circular orbit with radius R, and you shove it backward a bit, it will go into an elliptical orbit, and its speed will indeed get more as it comes further in. Its speed at perihelion, at radius r say, will be proportional to R/r (and not to the square).

But the speed of a planet in circular orbit with radius r will be (R/r)2 times the speed of a planet in circular orbit with radius R.

ah … Kepler's second law and conservation of momentum apply to the same orbit.

They don't apply if you start shoving extra energy in!

If a planet is in circular orbit with radius R, and you shove it backward a bit, it will go into an elliptical orbit, and its speed will indeed get more as it comes further in. Its speed at perihelion, at radius r say, will be proportional to R/r (and not to the square).

But the speed of a planet in circular orbit with radius r will be (R/r)2 times the speed of a planet in circular orbit with radius R.

Hi tiny-tim,

What you're saying is circular orbits obey my equation and an elliptical orbit got nothing to do with this equation, in another world, obey Kepler's 2nd law and conserve momentum?

The part make you confused was that suppose we want to change the planet into another stable circular orbit, say futher out. We have to 2 ways to do this:
1) push it outward and decrease its velocity.
2) increase its velocity and decrease its velocity much more when it reach outter orbit.

Staff: Mentor

tiny-tim is correct.

The gravitation force cancel out with the centrifugal force which equates into: Gm1m2/r^2 = m2v^2/r is a force balance, which maintains that the centripetal force is equal and opposite to the gravitational force. This is a static situation.

If one wishes to change an orbit, one must apply a force to work such that the gravitational potential energy changes. Applying a force means the system is not in equilibrium, but rather it is perturbed. Then we perturb the system again to obtain a circular orbit.

What you're saying is circular orbits obey my equation and an elliptical orbit got nothing to do with this equation, in another world, obey Kepler's 2nd law and conserve momentum?

Yes, your equation only works for the special case of a circular orbit. The full form that works for both circular and elliptical orbits is:
[tex]v= \sqrt{GM \left ( \frac{2}{r}- \frac{1}{a} \right )}[/tex]

Where "a" is the semi-major axis of the orbit, or the average orbital distance.
Note for a circular orbit a=r and it reduces to the simple form you used.

The part make you confused was that suppose we want to change the planet into another stable circular orbit, say futher out. We have to 2 ways to do this:
1) push it outward and decrease its velocity.
2) increase its velocity and decrease its velocity much more when it reach outter orbit.

For number 2, you increase its velocity to place it into the elliptical orbit which has its highest point at a distance equal to the radius of the new orbit you want. Then you increase its velocity to place it into a circular orbit. You need to increase the velocity because the object will lose velocity as it climbs away from the body that it is orbiting, and will being moving slower than the velocity needed to hold a circular orbit at the new altitude when it reaches it.

Now let's consider an over-exaggerated elliptical orbit so that from a far away point of view, it looks almost like linear. The motion one will see then is a bouncing object. How then one descibe the forces in this picture?

Now let's consider an over-exaggerated elliptical orbit so that from a far away point of view, it looks almost like linear. The motion one will see then is a bouncing object. How then one descibe the forces in this picture?

The cool part is we can interprete this picture as a pure elastic collision except the point of collision is not a physical collision. I'm trying to make this a model in microscopic scale. In theory sub atomic particle is so small that this model might suits well. I'll get back more on this.